K-diff Pairs in an Array

题目:

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an
integer pair (i, j), where i and j are both numbers in the array and their absolute
difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7

tips:难度其实不算大，主要是对于k=0的情况的处理，我吃了好几次亏，刚开始使用的是ArrayList来存储每个数，但是没办法判断或者说很难处理多个重复数据的情况，后来换用hashMap，用value来记录每个数出现的次数，这样就解决了最大的问题。

源码：

public class Solution {
public int findPairs(int[] nums, int k) {
if(nums==null || nums.length ==0 || k<0) return 0;
Map<Integer,Integer> map = new HashMap<>();
int count=0;
for(int n : nums) {
map.put(n,map.getOrDefault(n, 0)+1);
}
for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
if(k==0) {
if(entry.getValue()>=2) {
count++;
}
} else {
if(map.containsKey(entry.getKey()+k)) { //这里你+k或者-k都可以，结果没什么变化
count++;
}
}
}
return count;
}
}